## King 4 years ago Help.............Triangle ABC is a right triangle with angle ACB as the right angle. Angle ABC =60 degrees and AB=10. Let P be randomly chosen inside triangle ABC and extend BP to meet AC at D. What is the probability that BD>5*sqrt(2) Help.............Triangle ABC is a right triangle with angle ACB as the right angle. Angle ABC =60 degrees and AB=10. Let P be randomly chosen inside triangle ABC and extend BP to meet AC at D. What is the probability that BD>5*sqrt(2) @Mathematics

1. King

|dw:1320476146215:dw|

2. shubham_khandelwal

3. King

how?

4. shubham_khandelwal

5. King

how did u take as 15 and 45?

6. King

i never said that CBD is a right isosceles triangle

7. shubham_khandelwal

since we want hat BD> 5sqrt(2)........(I) we know that....10/sine(BDA) = BD/ sine(DAB) since.... sine(DAB) = 1/2 therefore..... BD = 5/ sine(BDA) put in equation no....(I) we get..... 5/sine(BDA)> 5*sqrt(2) yaha tak smjh me aa gaya kya sb kuch????

8. King

ya

9. shubham_khandelwal

so we get............ sin (BAD)<1/sqrt(2) or 180degre< BAD<135degree got it or not????

10. King

ok

11. King

From here i understood how to solve Thankx very very much fr ure help.......

12. shubham_khandelwal

now do u understand the figure that i have made....its the boundary line of the probability...that means in that figure if we move d to the right then we will get the desired probability....... understood????

13. King

And onee day hopefully like u i will be an IITIAN Yeah i understood THANX A LOT.......

14. shubham_khandelwal

listen tio the important fact........ur answer would be now....... area of triangle(CBD)/ area oof the triangle (DBA) this is the main thing....

15. King

ya kk

16. shubham_khandelwal

bye.....i'm leaving now...mujhe khana khane jana h aur bhi kuch kam h....i'll be back in two hrs now...do send me a request on facebook.......